我已经在我的网站的一项民意调查显示其单选按钮旁边的每个答案。当用户选择一个选项，提交，即时通过AJAX运行AA PHP脚本插入到表中的值或所选择的单选按钮。

我的阿贾克斯在运行，但目前插入0行的每一行，所以它不是拿起从单选按钮的值。任何帮助将是AP preciated。

HTML：

 ＆LT;形式ID =poll_form方法=后接收字符集=utf-8＆GT;
    ＆LT;输入类型=无线电名称=poll_option值=1ID =poll_option/＆GT;＆LT;标签='1'＆GT;＆安培; NBSP;艺术与LT; /标签＆gt;＆LT; BR /＆GT ;
    ＆LT;输入类型=无线电名称=poll_option值=2ID =poll_option/＆GT;＆LT;标签='2'＆GT;＆安培; NBSP;电影＆LT; /标签＆gt;＆LT; BR /＆GT ;
    ＆LT;输入类型=无线电名称=poll_option值=3ID =poll_option/＆GT;＆LT;标签='3'＆GT;＆安培; NBSP;游戏＆LT; /标签＆gt;＆LT; BR /＆GT ;
    ＆LT;输入类型=无线电名称=poll_option值=4ID =poll_option/＆GT;＆LT;标签='4'＆GT;＆安培; NBSP;音乐＆LT; /标签＆gt;＆LT; BR /＆GT ;
    ＆LT;输入类型=无线电名称=poll_option值=5ID =poll_option/＆GT;＆LT;标签='5'＆GT;＆安培; NBSP;体育和LT; /标签＆gt;＆LT; BR /＆GT ;
    ＆LT;输入类型=无线电名称=poll_option值=6ID =poll_option/＆GT;＆LT;标签='6'＆GT;＆安培; NBSP;电视＆LT; /标签＆gt;＆LT; BR /＆GT ;
    ＆LT;输入类型=提交值=投票和放大器; RARR; ID =submit_vote级=poll_btn/＆GT;
＆LT; /形式GT;
 

AJAX：

  $（＃submit_vote）。点击（函数（五）
    {
    VAR选项= $（'输入[类型=电台]：选中）VAL（）。
    $ optionID ==+ optionID;

    $阿贾克斯（{
        键入：POST，
        网址：ajax_submit_vote.php
        数据：{optionID：$ optionID}
    }）;
}）;
 

PHP：（缩短版）

 如果（$ _ SERVER ['REQUEST_METHOD'] ==POST）{

    //从提交的表单获得价值
    $选项= $ _ POST ['poll_option'];

    //插入到数据库
    $ insert_vote =INSERT INTO民意调查（USERIP，类别id）VALUES（'$ IP，$选项'）;
 

在此先感谢！

  $（＃submit_vote）。点击（函数（五）{

    $阿贾克斯（{
      键入：POST，
      网址：ajax_submit_vote.php
      数据：$（'＃poll_form'）序列化（）。
      成功：函数（响应）{}
    }）;

}）;
 

您应该然后在你的PHP脚本的POST变量poll_option访问。

I have a poll on my website which displays radio buttons next to each answer. When the user selects an option and submits, im running a a php script via ajax to insert the value or the selected radio button into a table.

My Ajax is running but is currently inserting a 0 row each row, so it's not picking up the value from the radio button. Any help would be appreciated.

HTML:

<form id="poll_form" method="post" accept-charset="utf-8">  
    <input type="radio" name="poll_option" value="1" id="poll_option" /><label for='1'>&nbsp;Arts</label><br />
    <input type="radio" name="poll_option" value="2" id="poll_option" /><label for='2'>&nbsp;Film</label><br />
    <input type="radio" name="poll_option" value="3" id="poll_option" /><label for='3'>&nbsp;Games</label><br />
    <input type="radio" name="poll_option" value="4" id="poll_option" /><label for='4'>&nbsp;Music</label><br />
    <input type="radio" name="poll_option" value="5" id="poll_option" /><label for='5'>&nbsp;Sports</label><br />
    <input type="radio" name="poll_option" value="6" id="poll_option" /><label for='6'>&nbsp;Television</label><br />    
    <input type="submit" value="Vote &rarr;" id="submit_vote" class="poll_btn"/> 
</form> 

AJAX:

    $("#submit_vote").click(function(e)
    { 
    var option=$('input[type="radio"]:checked').val();
    $optionID = "="+optionID;

    $.ajax({
        type: "POST",
        url: "ajax_submit_vote.php",
        data: {"optionID" : $optionID}
    });
});

PHP: (shortened version)

    if($_SERVER['REQUEST_METHOD'] == "POST"){

    //Get value from posted form
    $option = $_POST['poll_option'];

    //Insert into db
    $insert_vote = "INSERT into poll (userip,categoryid) VALUES ('$ip','$option')";

Thanks in advance!

$("#submit_vote").click(function(e){ 

    $.ajax( {
      type: "POST",
      url: "ajax_submit_vote.php",
      data: $('#poll_form').serialize(),
      success: function( response ) {}
    });

});

You should then have the POST variable "poll_option" accessible in your PHP script.